Lie bialgebroid

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In differential geometry, a field in mathematics, a Lie bialgebroid consists of two compatible Lie algebroids defined on dual vector bundles. Lie bialgebroids are the vector bundle version of Lie bialgebras.

Contents

Definition

Preliminary notions

A Lie algebroid consists of a bilinear skew-symmetric operation on the sections of a vector bundle over a smooth manifold , together with a vector bundle morphism subject to the Leibniz rule

and Jacobi identity

where are sections of and is a smooth function on .

The Lie bracket can be extended to multivector fields graded symmetric via the Leibniz rule

for homogeneous multivector fields .

The Lie algebroid differential is an -linear operator on the -forms of degree 1 subject to the Leibniz rule

for -forms and . It is uniquely characterized by the conditions

and

for functions on , -1-forms and sections of .

The definition

A Lie bialgebroid consists of two Lie algebroids and on the dual vector bundles and , subject to the compatibility

for all sections of . Here denotes the Lie algebroid differential of which also operates on the multivector fields .

Symmetry of the definition

It can be shown that the definition is symmetric in and , i.e. is a Lie bialgebroid if and only if is.

Examples

  1. A Lie bialgebra consists of two Lie algebras and on dual vector spaces and such that the Chevalley–Eilenberg differential is a derivation of the -bracket.
  2. A Poisson manifold gives naturally rise to a Lie bialgebroid on (with the commutator bracket of tangent vector fields) and (with the Lie bracket induced by the Poisson structure). The -differential is and the compatibility follows then from the Jacobi identity of the Schouten bracket.

Infinitesimal version of a Poisson groupoid

It is well known that the infinitesimal version of a Lie groupoid is a Lie algebroid (as a special case, the infinitesimal version of a Lie group is a Lie algebra). Therefore, one can ask which structures need to be differentiated in order to obtain a Lie bialgebroid.

Definition of Poisson groupoid

A Poisson groupoid is a Lie groupoid together with a Poisson structure on such that the graph of the multiplication map is coisotropic. An example of a Poisson-Lie groupoid is a Poisson-Lie group (where is a point). Another example is a symplectic groupoid (where the Poisson structure is non-degenerate on ).

Differentiation of the structure

Remember the construction of a Lie algebroid from a Lie groupoid. We take the -tangent fibers (or equivalently the -tangent fibers) and consider their vector bundle pulled back to the base manifold . A section of this vector bundle can be identified with a -invariant -vector field on which form a Lie algebra with respect to the commutator bracket on .

We thus take the Lie algebroid of the Poisson groupoid. It can be shown that the Poisson structure induces a fiber-linear Poisson structure on . Analogous to the construction of the cotangent Lie algebroid of a Poisson manifold there is a Lie algebroid structure on induced by this Poisson structure. Analogous to the Poisson manifold case one can show that and form a Lie bialgebroid.

Double of a Lie bialgebroid and superlanguage of Lie bialgebroids

For Lie bialgebras there is the notion of Manin triples, i.e. can be endowed with the structure of a Lie algebra such that and are subalgebras and contains the representation of on , vice versa. The sum structure is just

.

Courant algebroids

It turns out that the naive generalization to Lie algebroids does not give a Lie algebroid any more. Instead one has to modify either the Jacobi identity or violate the skew-symmetry and is thus lead to Courant algebroids. [1]

Superlanguage

The appropriate superlanguage of a Lie algebroid is , the supermanifold whose space of (super)functions are the -forms. On this space the Lie algebroid can be encoded via its Lie algebroid differential, which is just an odd vector field.

As a first guess the super-realization of a Lie bialgebroid should be . But unfortunately is not a differential, basically because is not a Lie algebroid. Instead using the larger N-graded manifold to which we can lift and as odd Hamiltonian vector fields, then their sum squares to iff is a Lie bialgebroid.

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References

  1. Z.-J. Liu, A. Weinstein and P. Xu: Manin triples for Lie bialgebroids, Journ. of diff. geom. vol. 45, pp. 547–574 (1997)